The velocity just before he touches the snow is given by "v_1"
Now,
using equation of motion for his free fall to snow
"v_1^2-u^2=2a_1s_1"
"u=0"
"a_1=9.8\\frac{m}{sec^2}"
"s_1=488ft=148.742m"
Hence,
"v_1=" "54\\frac{m}{sec}"
Now,
using equation of motion for decceleration by snow
"v^2-{v_1}^2=2a_2s_2"
"s_2=4.5ft=1.3716m"
"v=0"
Solving it we get,
"a_2=-1063\\frac{m}{sec^2}"
his average acceleration as he slowed to a stop is "1063\\frac{m}{sec^2}"
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